The value of K for the reaction 2 N a H C O 3 ( s ) ⇌ N a 2 C O 3 ( s ) + C O 2 ( g ) + H 2 O ( g ) has to be calculated.

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Answer 1

Question

Chapter 5, Problem 5I.16E

(a)

Interpretation Introduction

Interpretation:

The value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) has to be calculated.

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g)

The expression for equilibrium constant for the above reaction can be written as shown below,

K=(PCO2)(PH2O)

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

Composition(bar)NaHCO3Na2CO3CO2H2OInitial pressure−−00Change in pressure−−+x+xEquilibrium pressure−−xx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

K=(PCO2)(PH2O)=x2

Given that, the total pressure in the container is 7.68 bar.

x+x=7.68 barx=3.84 bar

Now, by plugging all data in the equilibrium constant expression, the value of K can be calculated.

K=(x)2=(3.84 atm)2K=14.74 atm2

Therefore, the value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) is 14.74 atm2.

(b)

Interpretation Introduction

Interpretation:

The equilibrium concentrations of CO2 and H2O has to be calculated.

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g)

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

Composition(bar)NaHCO3Na2CO3CO2H2OInitial pressure−−10Change in pressure−−+x+xEquilibrium pressure−−1+xx

Given that, the total pressure in the container is 7.68 bar.

1+x+x=7.68 barx=3.34 bar

Now, the equilibrium partial pressure of CO2 and H2O can be obtained as shown below.

PCO2=1+x=1+3.34=4.34 barPH2O=x=3.34 bar

The equilibrium concentration of CO2 and H2O can be obtained as shown below.

PV=nRTn(CO2)=PVRT=(4.34 bar)×(2.50 L)(8.3145×10−2L.bar.K−1.mol−1)(160+273)Kn(CO2)=0.30 mol[CO2]=0.30 mol2.50 L=0.12 Mn(H2O)=PVRT=(3.34 bar)×(2.50 L)(8.3145×10−2L.bar.K−1.mol−1)(160+273)Kn(H2O)=0.23 mol[H2O]=0.23 mol2.50 L=0.092 M

Therefore, the equilibrium concentration of CO2 and H2O is 0.12 M and 0.092 M respectively.

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Answer 2

Question

Chapter 5, Problem 5I.16E

(a)

Interpretation Introduction

Interpretation:

The value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) has to be calculated.

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g)

The expression for equilibrium constant for the above reaction can be written as shown below,

K=(PCO2)(PH2O)

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

Composition(bar)NaHCO3Na2CO3CO2H2OInitial pressure−−00Change in pressure−−+x+xEquilibrium pressure−−xx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

K=(PCO2)(PH2O)=x2

Given that, the total pressure in the container is 7.68 bar.

x+x=7.68 barx=3.84 bar

Now, by plugging all data in the equilibrium constant expression, the value of K can be calculated.

K=(x)2=(3.84 atm)2K=14.74 atm2

Therefore, the value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) is 14.74 atm2.

(b)

Interpretation Introduction

Interpretation:

The equilibrium concentrations of CO2 and H2O has to be calculated.

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g)

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

Composition(bar)NaHCO3Na2CO3CO2H2OInitial pressure−−10Change in pressure−−+x+xEquilibrium pressure−−1+xx

Given that, the total pressure in the container is 7.68 bar.

1+x+x=7.68 barx=3.34 bar

Now, the equilibrium partial pressure of CO2 and H2O can be obtained as shown below.

PCO2=1+x=1+3.34=4.34 barPH2O=x=3.34 bar

The equilibrium concentration of CO2 and H2O can be obtained as shown below.

PV=nRTn(CO2)=PVRT=(4.34 bar)×(2.50 L)(8.3145×10−2L.bar.K−1.mol−1)(160+273)Kn(CO2)=0.30 mol[CO2]=0.30 mol2.50 L=0.12 Mn(H2O)=PVRT=(3.34 bar)×(2.50 L)(8.3145×10−2L.bar.K−1.mol−1)(160+273)Kn(H2O)=0.23 mol[H2O]=0.23 mol2.50 L=0.092 M

Therefore, the equilibrium concentration of CO2 and H2O is 0.12 M and 0.092 M respectively.

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Answer 3

Question

Chapter 5, Problem 5I.16E

(a)

Interpretation Introduction

Interpretation:

The value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) has to be calculated.

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g)

The expression for equilibrium constant for the above reaction can be written as shown below,

K=(PCO2)(PH2O)

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

Composition(bar)NaHCO3Na2CO3CO2H2OInitial pressure−−00Change in pressure−−+x+xEquilibrium pressure−−xx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

K=(PCO2)(PH2O)=x2

Given that, the total pressure in the container is 7.68 bar.

x+x=7.68 barx=3.84 bar

Now, by plugging all data in the equilibrium constant expression, the value of K can be calculated.

K=(x)2=(3.84 atm)2K=14.74 atm2

Therefore, the value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) is 14.74 atm2.

(b)

Interpretation Introduction

Interpretation:

The equilibrium concentrations of CO2 and H2O has to be calculated.

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g)

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

Composition(bar)NaHCO3Na2CO3CO2H2OInitial pressure−−10Change in pressure−−+x+xEquilibrium pressure−−1+xx

Given that, the total pressure in the container is 7.68 bar.

1+x+x=7.68 barx=3.34 bar

Now, the equilibrium partial pressure of CO2 and H2O can be obtained as shown below.

PCO2=1+x=1+3.34=4.34 barPH2O=x=3.34 bar

The equilibrium concentration of CO2 and H2O can be obtained as shown below.

PV=nRTn(CO2)=PVRT=(4.34 bar)×(2.50 L)(8.3145×10−2L.bar.K−1.mol−1)(160+273)Kn(CO2)=0.30 mol[CO2]=0.30 mol2.50 L=0.12 Mn(H2O)=PVRT=(3.34 bar)×(2.50 L)(8.3145×10−2L.bar.K−1.mol−1)(160+273)Kn(H2O)=0.23 mol[H2O]=0.23 mol2.50 L=0.092 M

Therefore, the equilibrium concentration of CO2 and H2O is 0.12 M and 0.092 M respectively.

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Answer 4

Question

Chapter 5, Problem 5I.16E

(a)

Interpretation Introduction

Interpretation:

The value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) has to be calculated.

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g)

The expression for equilibrium constant for the above reaction can be written as shown below,

K=(PCO2)(PH2O)

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

Composition(bar)NaHCO3Na2CO3CO2H2OInitial pressure−−00Change in pressure−−+x+xEquilibrium pressure−−xx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

K=(PCO2)(PH2O)=x2

Given that, the total pressure in the container is 7.68 bar.

x+x=7.68 barx=3.84 bar

Now, by plugging all data in the equilibrium constant expression, the value of K can be calculated.

K=(x)2=(3.84 atm)2K=14.74 atm2

Therefore, the value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) is 14.74 atm2.

(b)

Interpretation Introduction

Interpretation:

The equilibrium concentrations of CO2 and H2O has to be calculated.

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g)

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

Composition(bar)NaHCO3Na2CO3CO2H2OInitial pressure−−10Change in pressure−−+x+xEquilibrium pressure−−1+xx

Given that, the total pressure in the container is 7.68 bar.

1+x+x=7.68 barx=3.34 bar

Now, the equilibrium partial pressure of CO2 and H2O can be obtained as shown below.

PCO2=1+x=1+3.34=4.34 barPH2O=x=3.34 bar

The equilibrium concentration of CO2 and H2O can be obtained as shown below.

PV=nRTn(CO2)=PVRT=(4.34 bar)×(2.50 L)(8.3145×10−2L.bar.K−1.mol−1)(160+273)Kn(CO2)=0.30 mol[CO2]=0.30 mol2.50 L=0.12 Mn(H2O)=PVRT=(3.34 bar)×(2.50 L)(8.3145×10−2L.bar.K−1.mol−1)(160+273)Kn(H2O)=0.23 mol[H2O]=0.23 mol2.50 L=0.092 M

Therefore, the equilibrium concentration of CO2 and H2O is 0.12 M and 0.092 M respectively.

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Answer 5

Chapter 5, Problem 5I.16E

(a)

Interpretation Introduction

Interpretation:

The value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) has to be calculated.

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g)

The expression for equilibrium constant for the above reaction can be written as shown below,

K=(PCO2)(PH2O)

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

Composition(bar)NaHCO3Na2CO3CO2H2OInitial pressure−−00Change in pressure−−+x+xEquilibrium pressure−−xx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

K=(PCO2)(PH2O)=x2

Given that, the total pressure in the container is 7.68 bar.

x+x=7.68 barx=3.84 bar

Now, by plugging all data in the equilibrium constant expression, the value of K can be calculated.

K=(x)2=(3.84 atm)2K=14.74 atm2

Therefore, the value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) is 14.74 atm2.

(b)

Interpretation Introduction

Interpretation:

The equilibrium concentrations of CO2 and H2O has to be calculated.

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g)

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

Composition(bar)NaHCO3Na2CO3CO2H2OInitial pressure−−10Change in pressure−−+x+xEquilibrium pressure−−1+xx

Given that, the total pressure in the container is 7.68 bar.

1+x+x=7.68 barx=3.34 bar

Now, the equilibrium partial pressure of CO2 and H2O can be obtained as shown below.

PCO2=1+x=1+3.34=4.34 barPH2O=x=3.34 bar

The equilibrium concentration of CO2 and H2O can be obtained as shown below.

PV=nRTn(CO2)=PVRT=(4.34 bar)×(2.50 L)(8.3145×10−2L.bar.K−1.mol−1)(160+273)Kn(CO2)=0.30 mol[CO2]=0.30 mol2.50 L=0.12 Mn(H2O)=PVRT=(3.34 bar)×(2.50 L)(8.3145×10−2L.bar.K−1.mol−1)(160+273)Kn(H2O)=0.23 mol[H2O]=0.23 mol2.50 L=0.092 M

Therefore, the equilibrium concentration of CO2 and H2O is 0.12 M and 0.092 M respectively.

Answer 6

Chapter 5, Problem 5I.16E

(a)

Interpretation Introduction

Interpretation:

The value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) has to be calculated.

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g)

The expression for equilibrium constant for the above reaction can be written as shown below,

K=(PCO2)(PH2O)

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

Composition(bar)NaHCO3Na2CO3CO2H2OInitial pressure−−00Change in pressure−−+x+xEquilibrium pressure−−xx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

K=(PCO2)(PH2O)=x2

Given that, the total pressure in the container is 7.68 bar.

x+x=7.68 barx=3.84 bar

Now, by plugging all data in the equilibrium constant expression, the value of K can be calculated.

K=(x)2=(3.84 atm)2K=14.74 atm2

Therefore, the value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) is 14.74 atm2.

Answer 7

Chapter 5, Problem 5I.16E

(a)

Interpretation Introduction

Interpretation:

The value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) has to be calculated.

Answer 8

(a)

Expert Solution

Explanation of Solution

The required reaction is given below.

2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g)

The expression for equilibrium constant for the above reaction can be written as shown below,

K=(PCO2)(PH2O)

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

Composition(bar)NaHCO3Na2CO3CO2H2OInitial pressure−−00Change in pressure−−+x+xEquilibrium pressure−−xx

Now, these values in the fourth row can be inserted in the equilibrium constant expression as shown below.

K=(PCO2)(PH2O)=x2

Given that, the total pressure in the container is 7.68 bar.

x+x=7.68 barx=3.84 bar

Now, by plugging all data in the equilibrium constant expression, the value of K can be calculated.

K=(x)2=(3.84 atm)2K=14.74 atm2

Therefore, the value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) is 14.74 atm2.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

The equilibrium concentrations of CO2 and H2O has to be calculated.

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g)

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

Composition(bar)NaHCO3Na2CO3CO2H2OInitial pressure−−10Change in pressure−−+x+xEquilibrium pressure−−1+xx

Given that, the total pressure in the container is 7.68 bar.

1+x+x=7.68 barx=3.34 bar

Now, the equilibrium partial pressure of CO2 and H2O can be obtained as shown below.

PCO2=1+x=1+3.34=4.34 barPH2O=x=3.34 bar

The equilibrium concentration of CO2 and H2O can be obtained as shown below.

PV=nRTn(CO2)=PVRT=(4.34 bar)×(2.50 L)(8.3145×10−2L.bar.K−1.mol−1)(160+273)Kn(CO2)=0.30 mol[CO2]=0.30 mol2.50 L=0.12 Mn(H2O)=PVRT=(3.34 bar)×(2.50 L)(8.3145×10−2L.bar.K−1.mol−1)(160+273)Kn(H2O)=0.23 mol[H2O]=0.23 mol2.50 L=0.092 M

Therefore, the equilibrium concentration of CO2 and H2O is 0.12 M and 0.092 M respectively.

Answer 13

(b)

Interpretation Introduction

Interpretation:

The equilibrium concentrations of CO2 and H2O has to be calculated.

Answer 14

(b)

Expert Solution

Explanation of Solution

The required reaction is given below.

2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g)

The change in pressure of any solid substance present in a reaction is always taken to be one.

An equilibrium table can be set up as given below.

Composition(bar)NaHCO3Na2CO3CO2H2OInitial pressure−−10Change in pressure−−+x+xEquilibrium pressure−−1+xx

Given that, the total pressure in the container is 7.68 bar.

1+x+x=7.68 barx=3.34 bar

Now, the equilibrium partial pressure of CO2 and H2O can be obtained as shown below.

PCO2=1+x=1+3.34=4.34 barPH2O=x=3.34 bar

The equilibrium concentration of CO2 and H2O can be obtained as shown below.

PV=nRTn(CO2)=PVRT=(4.34 bar)×(2.50 L)(8.3145×10−2L.bar.K−1.mol−1)(160+273)Kn(CO2)=0.30 mol[CO2]=0.30 mol2.50 L=0.12 Mn(H2O)=PVRT=(3.34 bar)×(2.50 L)(8.3145×10−2L.bar.K−1.mol−1)(160+273)Kn(H2O)=0.23 mol[H2O]=0.23 mol2.50 L=0.092 M

Therefore, the equilibrium concentration of CO2 and H2O is 0.12 M and 0.092 M respectively.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

Ch. 5 - Prob. 5A.1AST Ch. 5 - Prob. 5A.1BST Ch. 5 - Prob. 5A.2AST Ch. 5 - Prob. 5A.2BST Ch. 5 - Prob. 5A.3AST Ch. 5 - Prob. 5A.3BST Ch. 5 - Prob. 5A.1E Ch. 5 - Prob. 5A.2E Ch. 5 - Prob. 5A.3E Ch. 5 - Prob. 5A.4E

The value of K for the reaction 2 N a H C O 3 ( s ) ⇌ N a 2 C O 3 ( s ) + C O 2 ( g ) + H 2 O ( g ) has to be calculated.

Videos

Interpretation:

The value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) has to be calculated.

Explanation of Solution

Interpretation:

The equilibrium concentrations of CO2 and H2O has to be calculated.

Explanation of Solution

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Chapter 5 Solutions

The value of K for the reaction 2 N a H C O 3 ( s ) ⇌ N a 2 C O 3 ( s ) + C O 2 ( g ) + H 2 O ( g ) has to be calculated.

Videos

Interpretation: The value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) has to be calculated.

Explanation of Solution

Interpretation: The equilibrium concentrations of CO2 and H2O has to be calculated.

Explanation of Solution

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Chapter 5 Solutions

Interpretation:

The value of K for the reaction 2NaHCO3(s)⇌Na2CO3(s)+CO2(g)+H2O(g) has to be calculated.

Interpretation:

The equilibrium concentrations of CO2 and H2O has to be calculated.